UVa 116 - Unidirectional TSP

来源：互联网发布：字数统计英文编程编辑：程序博客网时间：2024/06/06 19:47

题目：一个二维的矩阵，从左向右找到一条路径，每次可以移动到右侧，右上或者右下的格子中，

要求找到一条，路径上的数字和最小的路径，输出路径（和相同时输出字典序最小的）。

分析：dp，动态三角形。因为要字典序最小逆序求解，记录输出即可，最优解取决于相邻的三个元素。

说明：注意输出格式。

#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;int maps[11][101];int smap[11][101];int fath[11][101];int main(){int n,m;while (~scanf("%d%d",&n,&m)) {for (int i = 1 ; i <= n ; ++ i)for (int j = 1 ; j <= m ; ++ j)scanf("%d",&maps[i][j]);memset(smap, 0, sizeof(smap));for (int i = m ; i >= 1 ; -- i)for (int j = 1 ; j <= n ; ++ j) {smap[j][i] = smap[j][i+1]+maps[j][i];fath[j][i] = j;if (j > 1 && smap[j][i] >= smap[j-1][i+1]+maps[j][i]) {smap[j][i] = smap[j-1][i+1]+maps[j][i];fath[j][i] = j-1;}if (j == n && smap[j][i] >= smap[1][i+1]+maps[j][i]) {smap[j][i] = smap[1][i+1]+maps[j][i];fath[j][i] = 1;}if (j < n && smap[j][i] > smap[j+1][i+1]+maps[j][i]) {smap[j][i] = smap[j+1][i+1]+maps[j][i];fath[j][i] = j+1;}if (j == 1 && smap[j][i] > smap[n][i+1]+maps[j][i]) {smap[j][i] = smap[n][i+1]+maps[j][i];fath[j][i] = n;}}int spa = 1;for (int i = 2 ; i <= n ; ++ i)if (smap[spa][1] > smap[i][1])spa = i;int min = smap[spa][1];for (int i = 1 ; i <= m ; ++ i) {if (i < m) printf("%d ",spa);else printf("%d\n%d\n",spa,min);spa = fath[spa][i];}}return 0;}

0 0